o Message expanded. Message read What is the Central Limit Theorem? posted by Jerome Tuttle , Jul 26, 2017, 6:38 AM Hi Class. I would like you to experiment with another population and sample, this time to explore the Central Limit Theorem. Suppose the population in column C has N=5,000 items. It is definitely not Normally distributed. (It is called a Pareto distribution, but you do not need to worry about the formulas I used in column C to create it.) It is skewed to the right, with many more small values than large values, which is common for incomes, housing prices, and insurance claims. The mean and SD are shown in cells K6 and K7. In column E, I take a sample of size n=50 from the population. Its sample mean xbar is in cell H5, which does not equal cell K6. Now suppose I take 100 samples (with replacement) of size n=50. I don't show you all 5,000 values, but I do show you in column H each sample mean xbar. When I graph the 100 sample means, the result is the sampling distribution of sample means. Page 285 lists the conclusions for the Central Limit Theorem. The spreadsheet uses Case 2 (original population is not Normally distributed). There are three conclusions: the sampling distribution of xbar is approximately ... ; the mean of all values of xbar equals ... ; and the SD of all values of xbar equals ... . These conclusions are supposed to work if I took ALL possible samples, but note that I only took 100 samples. So the conclusions are not going to work exactly here. However, would you say the three conclusions work approximately here? Would you say the three conclusions of Case 1 work approximately for the previous file, SamplingDistribution ? By the way, you can press F9 (recalculate), you get a new population and 100 new samples of size n=50. Words: 288 o What is a sampling distribution? posted by Jerome Tuttle , Jul 26, 2017, 6:32 AM o A sampling distribution (Section 6-4, page 272) can be a little difficult to get your arms around. Here is an example. Suppose the population in column C has N=1,000 items. It is approximately Normally distributed, with mean and SD shown in cells K6 and K7. In column E, I take a sample of size n=30 from the population. Its sample mean xbar is in cell H5, which does not equal cell K6. Now suppose I take 100 samples (with replacement) of size n=30. I don't show you all 3,000 values, but I do show you in column H each sample mean xbar. When I graph the 100 sample means, the result is the sampling distribution of sample means. I took 100 samples. If I took all possible samples of size n=30, the mean of all the xbars would equal the population mean, and the distribution would be exactly bell-shaped. Any comment? Any observation comparing the two graphs? Note that if you press F9 recalculate, you get a new population, new samples, new graphs, etc. Words: 171 Message expanded. Message read More soda cans o posted by Jerome Tuttle , Jul 26, 2017, 6:49 AM o Please read example 4, page 290-291. First, be sure you understand and can reproduce the mathematics, including the final step of calculating the probability of .0001. Now let's concentrate on the interpretation. With a population mean of 12 and a population SD of .11, there is a .0001 chance (1 in 10,000) of a 36 can sample of having a sample mean of 12.19. This is not a single 12.19 ounce can, but a mean of a 36 can sample at 12.19 - some cans are probably even more! It's possible, but extremely unlikely. Remember the Rare Event Rule says when an event occurs whose probability is less than 5%, that event is unusual and we should question the underlying assumptions. So that is what we should do here - we should question whether the population mean is really 12. We should reject that hypothesis - the population mean is probably more than 12. If the population mean is truly 12, there is a .0001 chance you can get a 36 can sample mean of 12.19. We will continue this kind of thinking in the remaining weeks of the course. Notice that I bolded population SD of .11. The manufacturer claims not only a mean of 12, but also a standard deviation of .11. That is a tiny SD, but it is part of the claim. That is part of the given information too, that students sometimes overlook. If I were a statistician employed by the soda manufacturer, I would urge them to figure out why they are (probably) over-filling their cans because they are losing money on each can. There is probably some sort of mechanical adjustment that is needed. Does this sort of thinking seem reasonable to you? Words: 264 Message expanded. Message read The advantage of mutual funds o posted by Jerome Tuttle , Jul 26, 2017, 6:44 AM . Here is a subtle point from the bottom of page 285: When working with an individual value from a normally distributed population, we use z = (x - ) / . But when working with a mean from a sample of sample size n, we have a distribution of sample means whose standard deviation is / n and we use z = (xbar - ) / ( / n ). So note that the variation among the sample means is much less than variation of the original data, as shown in the n. Suppose a Normally distributed population has = 100 and = 15. What is the probability that one x value is 105 or less? What is the probability that the mean of 30 values is 105 or less? A consequence is that someone who combines many individual items (such as a mutual fund which is a portfolio of many different stocks, or an insurance company which insures many different customers) will have less variation (per individual) in its results than just a single member can have. Any comment? In the online textbook, page 285 has a typo: the second equation has the square root of 2 in the denominator, but it should be the square root of n. Words: 187